Q.If $e^y (x+1) = 1$, show that $\frac{d^2 y}{dx^2} = \left(\frac{dy}{dx}\right)^2$.

Showing the 12 most recent of 50 on this concept.

1. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
   Q.If $\log\left(\sqrt{1+x^2}-x\right)=y\left(\sqrt{1+x^2}\right)$, then $(1+x^2)\dfrac{dy}{dx}+xy=$ (A) $0$ (B) $1$ (C) $2$ (D) $-1$

   ›Reveal solutionSolution

   Implicit differentiation of $\log(\sqrt{1+x^2}-x)=y\sqrt{1+x^2}$, using $(\sqrt{1+x^2}-x)(\sqrt{1+x^2}+x)=1$, collapses directly to the requested combination. Answer: $-1$.

   Concept and Intuition

   Writing $s=\sqrt{1+x^2}$ turns the relation into $\log(s-x)=ys$, a compact form whose derivative — after using the identity $s^2-x^2=1$ — telescopes into exactly the expression $(1+x^2)y'+xy$ asked for.

   Step-by-Step Solution

   1. Let $s=\sqrt{1+x^2}$; then $s'=\dfrac{x}{s}$ and $s^2-x^2=1\Rightarrow(s-x)(s+x)=1\Rightarrow\dfrac{1}{s-x}=s+x$.
   2. Given: $\log(s-x)=ys$. Differentiate both sides w.r.t. $x$: $\frac{s'-1}{s-x}=y's+ys'$ …

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2. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.If $x^x y^y = e^e$, then $\left(\dfrac{d^2y}{dx^2}\right)_{(e,e)} =$ (A) $\dfrac{1}{e}\left(\dfrac{dy}{dx}\right)_{(e,e)}$ (B) $\left(\dfrac{dy}{dx}\right)_{(e,e)} + \dfrac{1}{e}$ (C) $\left(\dfrac{dy}{dx}\right)_{(e,e)} - \dfrac{1}{e}$ (D) $e\left(\dfrac{dy}{dx}\right)_{(e,e)}$

   ›Reveal solutionSolution

   Logarithmic differentiation of $x^xy^y=e^e$ twice, evaluated at $(e,e)$, shows the second derivative equals $\frac1e$ times the first derivative there.

   Concept and Intuition

   Expressions like $x^x$ are best handled by taking logs first (since $\log(x^x)=x\log x$ is much easier to differentiate than $x^x$ directly). Implicit differentiation then relates $y'$ and $y''$ through the resulting equation.

   Step-by-Step Solution

   1. Take log: $x\log x + y\log y = \log(e^e) = e$ (a constant).
   2. Differentiate w.r.t. $x$: $(\log x+1) + (\log y+1)y' = 0$.
   3. Solve: $y' = -\dfrac{\log x+1}{\log y+1}$. At $(e,e)$: $\log e=1$, so $y'=-\dfrac{2}{2}=-1$.
   4. Differentiate the relation $(\log x+1)+(\log y+1)y'=0$ again w.r.t. $x$: $\dfrac1x + \dfrac{(y')^2}{y} + (\log y+1)y'' = 0$. …

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3. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
   Q.If $x^2+y^2=1$, then ______. (A) $y(y'')-4(y')^2+1=0$ (B) $y(y'')+(y')^2+1=0$ (C) $y(y'')-(y')^2-1=0$ (D) $y(y'')+2(y')^2+1=0$

   ›Reveal solutionSolution

   Differentiating the circle equation $x^2+y^2=1$ twice implicitly gives the differential equation $yy''+(y')^2+1=0$.

   Concept and Intuition

   Any implicit curve, when differentiated repeatedly with respect to $x$ treating $y$ as a function of $x$, yields a differential equation that the curve satisfies. Here we just need to differentiate twice and simplify.

   Step-by-Step Solution

   1. Start with $x^2+y^2=1$.
   2. Differentiate with respect to $x$: $2x+2yy'=0 \implies x+yy'=0$.
   3. Differentiate again with respect to $x$: $1 + (y')^2 + y y'' = 0$ (using the product rule on $yy'$, which gives $y'\cdot y' + y\cdot y''$).
   4. So the relation is $y y'' + (y')^2 + 1 = 0$. …

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4. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.m is the slope of a tangent to the curve $e^y = 1 + x^2$ at $x = 1$ then $m =$ (A) $\dfrac{2}{\log 2}$ (B) $\log 2$ (C) $2$ (D) $1$

   ›Reveal solutionSolution

   Implicit differentiation of $e^y=1+x^2$ gives slope $2x/(1+x^2)$, which is $1$ at $x=1$.

   Concept and Intuition

   This is a straightforward implicit differentiation: differentiate both sides with respect to $x$, treating $y$ as a function of $x$, then substitute the known relation back in to eliminate $e^y$.

   Step-by-Step Solution

   1. Differentiate $e^y = 1+x^2$ with respect to $x$: $e^y\dfrac{dy}{dx} = 2x$.
   2. So $\dfrac{dy}{dx} = \dfrac{2x}{e^y} = \dfrac{2x}{1+x^2}$ (substituting $e^y=1+x^2$ from the original equation). …

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5. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If $\sqrt{x-xy}+\sqrt{y-xy}=1$, then $\dfrac{dy}{dx}=$ (A) $-\sqrt{\dfrac{y-y^2}{x-x^2}}$ (B) $-\sqrt{\dfrac{1-y^2}{1-x^2}}$ (C) $-\sqrt{\dfrac{1-y}{1-x}}$ (D) $-\sqrt{\dfrac{x-y}{x+y}}$

   ›Reveal solutionSolution

   Squaring the constraint reveals that it forces $x+y=1$ identically, so $\dfrac{dy}{dx}=-1$ throughout; matching this against the options singles out $-\sqrt{\dfrac{y-y^2}{x-x^2}}$, since it equals $-1$ for every point satisfying $y=1-x$.

   Concept and Intuition

   Rather than blindly grinding through implicit differentiation of two square roots, it pays to first understand the curve itself. Squaring $\sqrt{x-xy}+\sqrt{y-xy}=1$ carefully (using $s=x+y,\,p=xy$) collapses to a perfect square equalling zero, revealing that the relation is nothing but the straight line $x+y=1$. Once we know that, $\dfrac{dy}{dx}=-1$ is immediate, and we just need to find which option reduces to $-1$ on this line.

   Step-by-Step Solution

   1. Square the given equation:

   $x(1-y)+y(1-x)+2\sqrt{xy(1-x)(1-y)} = 1$

   $x+y-2xy+2\sqrt{xy(1-x)(1-y)} = 1$

   2. Let $s=x+y,\ p=xy$. Then:

   $2\sqrt{p(1-x)(1-y)} = 1-s+2p$

   Note $(1-x)(1-y)=1-s+p$. Squaring again:

   $4p(1-s+p) = (1-s+2p)^2$

   Let $q=1-s$. Expanding both sides: LHS $=4pq+4p^2$; RHS $=q^2+4pq+4p^2$. So $0=q^2$, i.e. $q=0$, i.e. $s=1$.

   3. Hence $x+y=1$ is forced — the given relation is the line $y=1-x$ (restricted to the domain where the square roots are real, $0\le x,y\le1$).

   4. Differentiating $y=1-x$ directly: $\dfrac{dy}{dx}=-1$.

   5. Check which option gives $-1$ identically along $y=1-x$:

   • (A): $y-y^2=y(1-y)$. Substituting $y=1-x$: $y(1-y)=(1-x)\cdot x=x-x^2$. So the ratio is exactly $1$, and $-\sqrt1=-1$ for every $x$. ✓ …

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6. AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
   Q.If $y^{y^{y^{\cdot^{\cdot^{\cdot^{\infty}}}}}} = \log\{x + \log\{x + \cdots\}\}$, then $\dfrac{dy}{dx}$ at $x = e^2 - 2$, $y = \sqrt{2}$ equals _____ (A) $\dfrac{\log 2}{2\sqrt{2}(e^2 - 1)}$ (B) $\dfrac{1 - \log 2}{2\sqrt{2}(e^2 - 1)}$ (C) $\dfrac{\sqrt{2}(1 - \log 2)}{e^2 - 1}$ (D) $\dfrac{\log 2}{\sqrt{2}(e^2 - 1)}$

   ›Reveal solutionSolution

   Both sides define the same implicit quantity $u$ via a self-referential equation; differentiate each side's defining equation implicitly and combine using the chain rule. The answer is (B).

   Concept and Intuition

   The infinite power tower $y^{y^{y^{\cdots}}}=u$ satisfies the self-consistency equation $u=y^u$ (the tower "regenerates" itself). Likewise the infinite nested logarithm $\log\{x+\log\{x+\cdots\}\}=u$ satisfies $u=\log(x+u)$. Since the problem states these two quantities are equal (both equal to the same $u$), $u$ is implicitly a common function linking $x$ and $y$; differentiating each defining relation gives $du/dy$ and $du/dx$, and the chain rule combines them into $dy/dx$.

   Step-by-Step Solution

   1. Verify $u=2$ at the given point. Nested log: $u=\log(x+u)$ at $x=e^2-2$: try $u=2$: $\log(e^2-2+2)=\log(e^2)=2$ ✓. Tower: $u=y^u$ at $y=\sqrt2$: try $u=2$: $(\sqrt2)^2=2$ ✓. Both consistent with $u=2$.
   2. Differentiate the tower relation $u=y^u$ w.r.t. $y$. Take $\log$: $\log u = u\log y$. Differentiate: $\dfrac1u\dfrac{du}{dy} = \dfrac{du}{dy}\log y + \dfrac uy$ $\Rightarrow \left(\dfrac1u-\log y\right)\dfrac{du}{dy}=\dfrac uy \Rightarrow \dfrac{du}{dy}=\dfrac{u^2}{y(1-u\log y)}$. At $y=\sqrt2,\ u=2$: $\dfrac{du}{dy}=\dfrac{4}{\sqrt2(1-2\log\sqrt2)}=\dfrac{4}{\sqrt2(1-\ln2)}=\dfrac{2\sqrt2}{1-\ln2}$. …

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7. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.If $x^2+y^2=t+\dfrac{1}{t}$ and $x^4+y^4=t^2+\dfrac{1}{t^2}$, then $x^3 y\dfrac{dy}{dx}=$ (A) -1 (B) 0 (C) 1 (D) 2

   ›Reveal solutionSolution

   The two given relations force $x^2y^2=1$ (i.e. $xy$ is constant), from which $x^3y\,dy/dx=-1$.

   Concept and Intuition

   Rather than solving for $x,y$ in terms of $t$ explicitly, combine the two given equations algebraically (square the first, subtract the second) to eliminate $t$ entirely and land on a simple constant-product relation between $x$ and $y$.

   Step-by-Step Solution

   1. Square the first relation: $(x^2+y^2)^2=\left(t+\dfrac1t\right)^2=t^2+2+\dfrac{1}{t^2}$, i.e.

   $x^4+2x^2y^2+y^4=t^2+2+\frac{1}{t^2}$

   2. The second given relation is $x^4+y^4=t^2+\dfrac{1}{t^2}$.
   3. Subtract: $2x^2y^2 = \left(t^2+2+\dfrac1{t^2}\right)-\left(t^2+\dfrac1{t^2}\right)=2$, so $x^2y^2=1$.
   4. This means $xy=\pm1$, a constant independent of $t$. Differentiate $xy=\text{const}$ implicitly: …

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8. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.If $x^2 + y^2 = t - \dfrac{1}{t}$ and $x^4 + y^4 = t^2 + \dfrac{1}{t^2}$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{y}{x}$ (B) $\dfrac{y^2}{x^2}$ (C) $\sqrt{\dfrac{y}{x}}$ (D) $-\dfrac{y}{x}$

   ›Reveal solutionSolution

   Eliminating the parameter $t$ between the two given equations produces the direct relation $x^2y^2=-1$ between $x$ and $y$, whose implicit derivative is $-y/x$.

   Concept and Intuition

   When $x$ and $y$ are both linked to a parameter $t$ through two equations, differentiating each with respect to $t$ separately (and dividing) works, but it is often faster — and here it is exact — to first eliminate $t$ algebraically to get a direct $x$–$y$ relation, then differentiate that implicitly in the ordinary way.

   Step-by-Step Solution

   1. Square the first equation: $(x^2+y^2)^2=\left(t-\dfrac1t\right)^2=t^2-2+\dfrac1{t^2}$, i.e. $x^4+2x^2y^2+y^4=t^2+\dfrac1{t^2}-2$.
   2. The second equation says $x^4+y^4=t^2+\dfrac1{t^2}$. Substitute this in: $\left(t^2+\dfrac1{t^2}\right)+2x^2y^2=t^2+\dfrac1{t^2}-2$.
   3. This forces $2x^2y^2=-2\Rightarrow x^2y^2=-1$ — a $t$-free relation directly linking $x$ and $y$. …

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9. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If $\mathrm{Tan}^{-1}x^2 + \mathrm{Tan}^{-1}y^2 = \dfrac{\pi}{2}$, then $\left(\dfrac{dy}{dx}\right)_{(-1,2)} =$ (A) $0$ (B) $1$ (C) $\dfrac{1}{2}$ (D) $-\dfrac{1}{2}$

   ›Reveal solutionSolution

   Reducing to $y^2=x^{-2}$ gives $\dfrac{dy}{dx}=-\dfrac{1}{x^3 y}$, which at $(-1,2)$ equals $\dfrac12$.

   Concept and Intuition

   If $\mathrm{Tan}^{-1}a+\mathrm{Tan}^{-1}b=\dfrac{\pi}{2}$ with $a,b>0$, then $\mathrm{Tan}^{-1}b=\dfrac{\pi}{2}-\mathrm{Tan}^{-1}a=\mathrm{Cot}^{-1}a$, so $b=\dfrac{1}{a}$. Applying this to $a=x^2,\ b=y^2$ collapses the relation into an algebraic one.

   Step-by-Step Solution

   1. From $\mathrm{Tan}^{-1}x^2+\mathrm{Tan}^{-1}y^2=\dfrac{\pi}{2}$ we get $y^2=\dfrac{1}{x^2}=x^{-2}$.
   2. Differentiate: $2y\,\dfrac{dy}{dx}=-2x^{-3}$.
   3. Hence $\dfrac{dy}{dx}=-\dfrac{1}{x^3 y}$. …

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10. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
    Q.If $x^2 + y^2 + \sin y = 4$, then the value of $\dfrac{d^2y}{dx^2}$ at $x = -2$ is (A) $-30$ (B) $-34$ (C) $-32$ (D) $-18$

    ›Reveal solutionSolution

    Implicit differentiation of $x^2+y^2+\sin y=4$ twice, using $y(-2)=0$ and $y'(-2)=4$, gives $y''(-2)=-34$.

    Concept and Intuition

    For an implicitly-defined curve, we differentiate the whole equation with respect to $x$ (treating $y$ as a function of $x$ and applying the chain rule to every $y$-term), solve for $y'$, then differentiate the resulting equation again to get $y''$. The key first step is always finding the actual point $(x_0,y_0)$ on the curve, since $y'$ and $y''$ are evaluated there.

    Step-by-Step Solution

    1. Find $y$ at $x=-2$: substituting $x=-2$ into $x^2+y^2+\sin y=4$: $4+y^2+\sin y=4\Rightarrow y^2+\sin y=0$. Clearly $y=0$ satisfies this (and is the relevant branch), so $y(-2)=0$.
    2. First derivative: differentiate $x^2+y^2+\sin y=4$ w.r.t. $x$:

    $2x+2y\,y'+\cos y\cdot y'=0 \implies y'(2y+\cos y)=-2x \implies y'=\frac{-2x}{2y+\cos y}.$

    At $(x,y)=(-2,0)$: $y'=\dfrac{-2(-2)}{2(0)+\cos0}=\dfrac{4}{1}=4$.

    3. Second derivative: differentiate $2x+2y\,y'+\cos y\cdot y'=0$ again w.r.t. $x$, using the product rule on both $2y\,y'$ and $\cos y \cdot y'$: …

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11. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
    Q.If $y = \log_y x$, then $\dfrac{dy}{dx} =$ ______ (A) $\dfrac{1}{x\log y}$ (B) $\dfrac{\log y}{x(1+\log y)}$ (C) $\dfrac{1}{x(1+\log y)}$ (D) $\dfrac{1}{1+\log y}$

    ›Reveal solutionSolution

    Rewriting $y=\log_y x$ as $y\ln y=\ln x$ and differentiating implicitly gives $\dfrac{dy}{dx}=\dfrac{1}{x(1+\log y)}$.

    Concept and Intuition

    $\log_y x$ means "logarithm of $x$ to base $y$", i.e. $\dfrac{\ln x}{\ln y}$. So the given relation $y=\log_y x$ really means $y=\dfrac{\ln x}{\ln y}$, or equivalently $y\ln y=\ln x$ — a cleaner form to differentiate implicitly, since it avoids a quotient with $y$ in both places.

    Step-by-Step Solution

    1. $y=\log_y x=\dfrac{\ln x}{\ln y} \Rightarrow y\ln y=\ln x$.
    2. Differentiate both sides with respect to $x$, treating $y$ as a function of $x$:

    $\frac{d}{dx}(y\ln y)=\frac{d}{dx}(\ln x)$

    3. LHS (product rule): $\dfrac{dy}{dx}\ln y+y\cdot\dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{dy}{dx}(\ln y+1)$.
    4. RHS: $\dfrac1x$.
    5. So $\dfrac{dy}{dx}(\ln y+1)=\dfrac1x \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{x(1+\ln y)}=\dfrac{1}{x(1+\log y)}$. …

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12. AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
    Q.If $y = \sqrt{x + \sqrt{x + \sqrt{x + \cdots \infty}}}$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{1}{y}$ (B) $\dfrac{1}{x}$ (C) $\dfrac{1}{2x-1}$ (D) $\dfrac{1}{2y-1}$

    ›Reveal solutionSolution

    The infinite nested radical satisfies $y^2=x+y$ (self-similarity), which is then differentiated implicitly.

    Concept and Intuition

    An infinitely repeating nested expression under a radical satisfies a self-referential equation: the whole expression $y$ equals the same structure with $x+y$ under the first radical (since removing the outermost layer just reproduces $y$ again).

    Step-by-Step Solution

    1. $y=\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}=\sqrt{x+y}$ (the inner infinite tail is again $y$).
    2. Square both sides: $y^2=x+y$.
    3. Differentiate implicitly with respect to $x$: $2y\dfrac{dy}{dx}=1+\dfrac{dy}{dx}$.
    4. Collect: $\dfrac{dy}{dx}(2y-1)=1\Rightarrow\dfrac{dy}{dx}=\dfrac1{2y-1}$. …

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